I'm trying to compute $$\int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx$$ i.e. the Fourier transform of $x\mapsto \frac{\sinh(kx)}{\sinh(x)}$, where $0<k<1$ is fixed.
But I'm having trouble with it.
Motivation: I'm trying to derive an expression for the solution of the Dirichlet problem $\Delta u = 0$ on the strip $[0,1]\times \mathbb R$ with values on the boundary $f_0, f_1$ (assuming necessary niceness conditions for all functions involved).
For this I took the Fourier transform of the solution $u$ and got a formula for $\hat u$ in terms of $\hat{ f_0}, \hat{ f_1}$ and now I want to transform it back. Doing this led (more or less) to the integral expression: $u(x,y) = \int \frac{\sinh(kx)}{\sinh(k)}e^{-iky} \hat f(k) \ dk$.
Now I'm trying to apply the product formula $\int f \hat g = \int \hat f g$ to get everything in terms of $f$. This is why I'm interested in computing the above integral.
My attempt: (which led nowhere, so you may actually ignore everything below)
I think it should be possible using residues. For this I thought of the path having the following components:
$$[-R,R], \ [R,R+i\pi], \ [R+i\pi, \delta + i\pi],$$
$$ \ \text{semicircle from $\delta + i\pi$ to $-\delta + i\pi$ below $i\pi$},$$
$$[-\delta + i\pi, -R + i\pi], \ [-R+i\pi, -R]$$
with the intention of letting $\delta \to 0$ eventually.
The integrals over the vertical components will vanish for $R\to\infty$, so the integral over the path then becomes
\begin{align}
0 &= \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx + \left(\int_{\infty}^{\delta} + \int_{-\delta}^{-\infty}\right) \frac{\sinh(k(x+i\pi))}{\sinh(x+i\pi)}e^{-i\omega (x+i\pi)} \ dx \\ & \qquad + \int_{\text{semicircle}} \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx
\end{align}
Using $\sinh(a+ib) = \sinh(a)\cos(b) + i \cosh(a)\sin(b)$ for real $a,b$, we get
\begin{align}
0 &= \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \\ &\qquad + \left(\int_{\delta}^{\infty} + \int_{-\infty}^{-\delta}\right) \frac{\sinh(kx)\cos(k\pi) + i \cosh(kx)\sin(k\pi)}{\sinh(x)}e^{-i\omega x}e^{\omega \pi} \ dx \\ & \qquad + \int_{\text{semicircle}} \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx
\end{align}
The integral over the semicircle should go to $$(-\pi i) \ \mathrm{Res}_{x = \pi i}\left(\frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x}\right) = -\pi \sin(k\pi)e^{\omega \pi}$$
as $\delta \to 0$. Therefore
\begin{align}
\pi \sin(k\pi) e^{\omega \pi} &= \int_{-\infty}^\infty \frac{\sinh(kx)(1+\cos(k\pi)e^{\omega \pi}) + i \cosh(kx) \sin(x\pi)e^{\omega \pi}} {\sinh(x)} e^{-i\omega x} \ dx \\
&= (1+\cos(k\pi)e^{\omega \pi}) \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \\
& \qquad + i \sin(k\pi)e^{\omega \pi}\int_{-\infty}^\infty \frac{\cosh(kx)}{\sinh(x)}e^{-i\omega x} \ dx
\end{align}
I don't see whether this has brought me any closer to my goal?
Best Answer
The result is doable by method of residues. We complete the integration path by the arc crossing from $+\infty$ to $-\infty$ over the upper-half complex plane. Then $$ \begin{eqnarray} \mathcal{F}(\omega, \kappa) &=& \int_{-\infty}^\infty \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \mathrm{d} x = 2 \pi i \sum_{n=1}^\infty \operatorname{Res}_{x = i \pi n} \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \\ &=& \sum_{n=1}^\infty 2 \pi (-1)^{n-1} \mathrm{e}^{-\omega \pi n} \sin(\pi \kappa n) = \frac{2 \pi e^{\pi \omega } \sin (\pi \kappa )}{2 e^{\pi \omega } \cos (\pi \kappa )+e^{2 \pi \omega }+1} \\ &=& \frac{\pi \sin (\pi \kappa )}{\cos (\pi \kappa )+\cosh\left( \pi \omega \right)} \end{eqnarray} $$